Chapter 4
Bruce Kim
October 11, 2016
- This question should be answered using the Weekly data set, which is part of the ISLR package. This data is similar in nature to the Smarket data from this chapter’s lab, except that it contains 1, 089 weekly returns for 21 years, from the beginning of 1990 to the end of
data <- Weekly
attach(data)
itrain <- createDataPartition(y = Direction, p = 0.75, list = FALSE)
train <- data[itrain,]
test <- data[-itrain,]- Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
Given the pair plot, the volume has been exponentially increased. It seems to come from the dot-com bubble. Other than that, there is no clear sign on the relationship between variables.
par(mar = c(1,1,1,1))
pairs(train)
Just for fun, let’s see whether the stock return shows the normal curve. It clearly shows the normal curve.
library(gridExtra)
lag1 <- qplot(x = Lag1, data = train, main = "Histogram of Lag 1")
lag2 <- qplot(x = Lag2, data = train, main = "Histogram of Lag 2")
lag3 <- qplot(x = Lag3, data = train, main = "Histogram of Lag 3")
lag4 <- qplot(x = Lag4, data = train, main = "Histogram of Lag 4")
lag5 <- qplot(x = Lag5, data = train, main = "Histogram of Lag 5")
Today <- qplot(x = Today, data = train, main = "Histogram of Lag 6")
grid.arrange(lag1,lag2,lag3,lag4,lag5,Today, nrow = 3, ncol = 2)## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
- Use the full data set to perform a logistic regression with Direction as the response and the five lag variables plus Volume as predictors. Use the summary function to print the results. Do any of the predictors appear to be statistically significant? If so, which ones?
ctrl <- trainControl(method = "repeatedcv", number = 10)
glm.fit <- train(Direction ~ . -Year -Today,data = train, trControl = ctrl, method = "glm",
family = "binomial")
summary(glm.fit)##
## Call:
## NULL
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.8109 -1.2578 0.9954 1.0806 1.3938
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.30620 0.09930 3.084 0.00204 **
## Lag1 -0.05166 0.03083 -1.676 0.09376 .
## Lag2 0.02653 0.03033 0.875 0.38173
## Lag3 -0.02157 0.03139 -0.687 0.49190
## Lag4 -0.01613 0.03113 -0.518 0.60429
## Lag5 -0.04326 0.03043 -1.422 0.15509
## Volume -0.04179 0.04270 -0.979 0.32774
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1122.4 on 816 degrees of freedom
## Residual deviance: 1115.1 on 810 degrees of freedom
## AIC: 1129.1
##
## Number of Fisher Scoring iterations: 4- Compute the confusion matrix and overall fraction of correct predictions. Explain what the confusion matrix is telling you about the types of mistakes made by logistic regression.
pred <- predict(glm.fit, test)
cm.glm <- confusionMatrix(test$Direction,pred)
cm.glm## Confusion Matrix and Statistics
##
## Reference
## Prediction Down Up
## Down 12 109
## Up 22 129
##
## Accuracy : 0.5184
## 95% CI : (0.4572, 0.5791)
## No Information Rate : 0.875
## P-Value [Acc > NIR] : 1
##
## Kappa : -0.0501
## Mcnemar's Test P-Value : 5.741e-14
##
## Sensitivity : 0.35294
## Specificity : 0.54202
## Pos Pred Value : 0.09917
## Neg Pred Value : 0.85430
## Prevalence : 0.12500
## Detection Rate : 0.04412
## Detection Prevalence : 0.44485
## Balanced Accuracy : 0.44748
##
## 'Positive' Class : Down
## - Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
glm.fit <- train(Direction ~ Lag2, data = train, trControl = ctrl, method = "glm", family = "binomial")
pred <- predict(glm.fit, test)
cm.glm <- confusionMatrix(test$Direction, pred)
cm.glm## Confusion Matrix and Statistics
##
## Reference
## Prediction Down Up
## Down 1 120
## Up 1 150
##
## Accuracy : 0.5551
## 95% CI : (0.4939, 0.6152)
## No Information Rate : 0.9926
## P-Value [Acc > NIR] : 1
##
## Kappa : 0.0018
## Mcnemar's Test P-Value : <2e-16
##
## Sensitivity : 0.500000
## Specificity : 0.555556
## Pos Pred Value : 0.008264
## Neg Pred Value : 0.993377
## Prevalence : 0.007353
## Detection Rate : 0.003676
## Detection Prevalence : 0.444853
## Balanced Accuracy : 0.527778
##
## 'Positive' Class : Down
## - Repeat (d) using LDA.
lda.fit <- train(Direction ~ Lag2, data = train, trControl = ctrl, method = "lda")## Loading required package: MASSpred.lda <- predict(lda.fit, test)
cm.lda <- confusionMatrix(test$Direction, pred.lda)
cm.lda## Confusion Matrix and Statistics
##
## Reference
## Prediction Down Up
## Down 1 120
## Up 1 150
##
## Accuracy : 0.5551
## 95% CI : (0.4939, 0.6152)
## No Information Rate : 0.9926
## P-Value [Acc > NIR] : 1
##
## Kappa : 0.0018
## Mcnemar's Test P-Value : <2e-16
##
## Sensitivity : 0.500000
## Specificity : 0.555556
## Pos Pred Value : 0.008264
## Neg Pred Value : 0.993377
## Prevalence : 0.007353
## Detection Rate : 0.003676
## Detection Prevalence : 0.444853
## Balanced Accuracy : 0.527778
##
## 'Positive' Class : Down
## - Repeat (d) using QDA.
qda.fit <- train(Direction ~ Lag2, data = train, trControl = ctrl, method = "qda")
pred.qda <- predict(qda.fit, test)
cm.qda <- confusionMatrix(test$Direction, pred.qda)
cm.qda## Confusion Matrix and Statistics
##
## Reference
## Prediction Down Up
## Down 0 121
## Up 0 151
##
## Accuracy : 0.5551
## 95% CI : (0.4939, 0.6152)
## No Information Rate : 1
## P-Value [Acc > NIR] : 1
##
## Kappa : 0
## Mcnemar's Test P-Value : <2e-16
##
## Sensitivity : NA
## Specificity : 0.5551
## Pos Pred Value : NA
## Neg Pred Value : NA
## Prevalence : 0.0000
## Detection Rate : 0.0000
## Detection Prevalence : 0.4449
## Balanced Accuracy : NA
##
## 'Positive' Class : Down
## - Repeat (d) using KNN with K = 1.
knn.fit <- train(Direction ~ Lag2, data = train, trControl = ctrl, method = "knn", tuneLength = 1)
pred.knn <- predict(knn.fit, test)
cm.knn <- confusionMatrix(test$Direction, pred.knn)
cm.knn## Confusion Matrix and Statistics
##
## Reference
## Prediction Down Up
## Down 50 71
## Up 57 94
##
## Accuracy : 0.5294
## 95% CI : (0.4682, 0.59)
## No Information Rate : 0.6066
## P-Value [Acc > NIR] : 0.9959
##
## Kappa : 0.0362
## Mcnemar's Test P-Value : 0.2505
##
## Sensitivity : 0.4673
## Specificity : 0.5697
## Pos Pred Value : 0.4132
## Neg Pred Value : 0.6225
## Prevalence : 0.3934
## Detection Rate : 0.1838
## Detection Prevalence : 0.4449
## Balanced Accuracy : 0.5185
##
## 'Positive' Class : Down
## - Which of these methods appears to provide the best results on this data? All model is similar. Don’t invest with this model. It’s highway to Yodan River.
detach(data)
library(knitr)
compr <- list(cm.glm$overall,cm.lda$overall,cm.qda$overall,cm.knn$overall)
compr <- data.frame(do.call(rbind, compr), row.names = c("glm", "lda", "qda", "knn"))
kable(compr)| Accuracy | Kappa | AccuracyLower | AccuracyUpper | AccuracyNull | AccuracyPValue | McnemarPValue | |
|---|---|---|---|---|---|---|---|
| glm | 0.5551471 | 0.0018197 | 0.4939236 | 0.6151558 | 0.9926471 | 1.0000000 | 0.000000 |
| lda | 0.5551471 | 0.0018197 | 0.4939236 | 0.6151558 | 0.9926471 | 1.0000000 | 0.000000 |
| qda | 0.5551471 | 0.0000000 | 0.4939236 | 0.6151558 | 1.0000000 | 1.0000000 | 0.000000 |
| knn | 0.5294118 | 0.0361553 | 0.4682146 | 0.5899612 | 0.6066176 | 0.9959414 | 0.250536 |
- In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.
- Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
library(dplyr)##
## Attaching package: 'dplyr'## The following object is masked from 'package:MASS':
##
## select## The following object is masked from 'package:gridExtra':
##
## combine## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, uniondata <- Auto
attach(data)## The following object is masked from package:ggplot2:
##
## mpgmpg <- as.numeric(mpg)
acceleration <- as.numeric(acceleration)
displacement <- as.numeric(displacement)
horsepower <- as.numeric(horsepower)
origin <- as.factor(origin)
weight <- as.numeric(weight)
year <- as.factor(year)
cylinders <- as.factor(cylinders)
data <- mutate(data, mpg01 = ifelse(mpg < median(mpg), 0, 1))
data$mpg01 <- as.factor(data$mpg01)- Explore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.
attach(data)## The following objects are masked _by_ .GlobalEnv:
##
## acceleration, cylinders, displacement, horsepower, mpg,
## origin, weight, year## The following objects are masked from data (pos = 3):
##
## acceleration, cylinders, displacement, horsepower, mpg, name,
## origin, weight, year## The following object is masked from package:ggplot2:
##
## mpgset.seed(1354)
iTrain <- createDataPartition(y = mpg01, p = 0.75, list = FALSE)
Train <- data[iTrain, ]
Test <- data[-iTrain, ]
detach(data)First, all qualitative variables shows high correlation with mpg. Also it’s interesting that three variables, displacement, horsepower, and weight, show the nonlinearity as respect to mpg. But, each explanatory variables shows correlation with each, called corlinearity.
attach(Train)## The following objects are masked _by_ .GlobalEnv:
##
## acceleration, cylinders, displacement, horsepower, mpg,
## origin, weight, year## The following objects are masked from data:
##
## acceleration, cylinders, displacement, horsepower, mpg, name,
## origin, weight, year## The following object is masked from package:ggplot2:
##
## mpgDumforPairs <- Train[,-c(2, 7, 8,9, 10)]
pairs(DumforPairs)
Second, those are the box plot as respect to mpg01
b.q.1 <- qplot(x = mpg01, y = displacement, data = Train, geom = "boxplot",
main = "Displacement")
b.q.2 <- qplot(x = mpg01, y = horsepower, data = Train, geom = "boxplot",
main = "Horsepower")
b.q.3 <- qplot(x = mpg01, y = weight, data = Train, geom = "boxplot",
main = "Weight")
b.q.4 <- qplot(x = mpg01, y = acceleration, data = Train, geom = "boxplot",
main = "Acceleration")
grid.arrange(b.q.1, b.q.2, b.q.3, b.q.4, ncol = 2)
Unlike other variables, Acceleration shows any significant difference as respect to the fuel efficiency, e.g. mpg.
- Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
ctrl <- trainControl(method = "repeatedcv")
names(Train)## [1] "mpg" "cylinders" "displacement" "horsepower"
## [5] "weight" "acceleration" "year" "origin"
## [9] "name" "mpg01"fit.LDA <- train(mpg01 ~. - name - mpg, data = Train, trControl = ctrl, method = "lda")
pred.LDA <- predict(fit.LDA, Test)
cm.LDA <- confusionMatrix(pred.LDA, Test$mpg01)
1 - cm.LDA$overall[[1]]## [1] 0.09183673d1 <- data.frame(cm.LDA$byClass)- Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
ctrl <- trainControl(method = "repeatedcv")
fit.QDA <- train(mpg01 ~. - name - mpg, data = Train, trControl = ctrl, method = "qda")
pred.QDA <- predict(fit.QDA, Test)
cm.QDA <- confusionMatrix(pred.QDA, Test$mpg01)
1 - cm.QDA$overall[[1]]## [1] 0.1122449d2 <- data.frame(cm.QDA$byClass)- Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
ctrl <- trainControl(method = "repeatedcv")
fit.glm <- train(mpg01 ~. - name - mpg, data = Train, trControl = ctrl, method = "glm", family = "binomial")
pred.glm <- predict(fit.glm, Test)
cm.glm <- confusionMatrix(pred.glm, Test$mpg01)
1 - cm.glm$overall[[1]]## [1] 0.122449- Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
ctrl <- trainControl(method = "repeatedcv")
fit.knn <- train(mpg01 ~. - name - mpg, data = Train, trControl = ctrl, method = "knn", preProcess = c("center", "scale"), tuneLength = 20)
pred.knn <- predict(fit.knn, Test)
cm.knn <- confusionMatrix(pred.knn, Test$mpg01)
result <- fit.knn$results
kable(result[1:5,])| k | Accuracy | Kappa | AccuracySD | KappaSD |
|---|---|---|---|---|
| 5 | 0.9050411 | 0.8097348 | 0.0545497 | 0.1095113 |
| 7 | 0.9052791 | 0.8100772 | 0.0652501 | 0.1313458 |
| 9 | 0.9050411 | 0.8096010 | 0.0613871 | 0.1236705 |
| 11 | 0.9017077 | 0.8029344 | 0.0625845 | 0.1260201 |
| 13 | 0.9017077 | 0.8029344 | 0.0625845 | 0.1260201 |
After preprocessing the data, k=7 shows the highest accuracy.
d3 <- data.frame(cm.glm$byClass)
d4 <- data.frame(cm.knn$byClass)
D <- cbind(d1,d2,d3,d4); colnames(D) <- c("LDA", "QDA", "GLM", "KNN")
ac <- cbind(cm.LDA$overall[1], cm.QDA$overall[1], cm.glm$overall[1], cm.knn$overall[1])
colnames(ac) <- c("LDA", "QDA", "GLM", "KNN")
D <- rbind(ac,D)
kable(D[1:5, ])| LDA | QDA | GLM | KNN | |
|---|---|---|---|---|
| Accuracy | 0.9081633 | 0.8877551 | 0.8775510 | 0.9081633 |
| Sensitivity | 0.8367347 | 0.8571429 | 0.8163265 | 0.8367347 |
| Specificity | 0.9795918 | 0.9183673 | 0.9387755 | 0.9795918 |
| Pos Pred Value | 0.9761905 | 0.9130435 | 0.9302326 | 0.9761905 |
| Neg Pred Value | 0.8571429 | 0.8653846 | 0.8363636 | 0.8571429 |
| First, both LDA a | nd KNN shows | the highest | overall acc | uracy. For specificity, LDA ranked the first followed by KNN. If the purpose for the prediction is to find the car having high fuel efficiency, it is better off chosing LDA as a predictive model. |
- Using the Boston data set, fit classification models in order to predict whether a given suburb has a crime rate above or below the median. Explore logistic regression, LDA, and KNN models using various subsets of the predictors. Describe your findings.
library(MASS)
data <- Boston
kable(head(Boston))| crim | zn | indus | chas | nox | rm | age | dis | rad | tax | ptratio | black | lstat | medv |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 0.00632 | 18 | 2.31 | 0 | 0.538 | 6.575 | 65.2 | 4.0900 | 1 | 296 | 15.3 | 396.90 | 4.98 | 24.0 |
| 0.02731 | 0 | 7.07 | 0 | 0.469 | 6.421 | 78.9 | 4.9671 | 2 | 242 | 17.8 | 396.90 | 9.14 | 21.6 |
| 0.02729 | 0 | 7.07 | 0 | 0.469 | 7.185 | 61.1 | 4.9671 | 2 | 242 | 17.8 | 392.83 | 4.03 | 34.7 |
| 0.03237 | 0 | 2.18 | 0 | 0.458 | 6.998 | 45.8 | 6.0622 | 3 | 222 | 18.7 | 394.63 | 2.94 | 33.4 |
| 0.06905 | 0 | 2.18 | 0 | 0.458 | 7.147 | 54.2 | 6.0622 | 3 | 222 | 18.7 | 396.90 | 5.33 | 36.2 |
| 0.02985 | 0 | 2.18 | 0 | 0.458 | 6.430 | 58.7 | 6.0622 | 3 | 222 | 18.7 | 394.12 | 5.21 | 28.7 |
It turned out the classes of varaibles in the data are quite messy. For further analysis, I chagned the classes of variables according to the code book in help page.
The data, boston, has 14 variables along with 506 observations. We would predict the degree of high crime rate with ohter 13 variables. First, we need to manipulate, then split the data for further analysis.
set.seed(34596)
iTrain <- createDataPartition(y = data$Mcrim, p = 0.75, list = FALSE)
Train <- data[iTrain, ]
Test <- data[-iTrain, ]Here is the summary of the data.
summary(data)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv Mcrim
## Min. : 1.73 Min. : 5.00 0:253
## 1st Qu.: 6.95 1st Qu.:17.02 1:253
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00The interesting factor in the Boston data is chas variable. The variable indicates whether the town bounds Charls river. Does the closeness to Charls river show the relationship with crime rate? With the table, we can see whether the relation exists. The per centage of crime goes above median if the town is far from the river is 49%. And that of crime goes above median if the town bounds the river is 62.5%.
attach(Train)
table(chas,Mcrim)## Mcrim
## chas 0 1
## 0 181 175
## 1 9 15Next, let’s look at the box plot as respect to the degree of crime, Mcrim.
q1 <- qplot(x = Mcrim, y = log(zn)+1, data = Train, geom = "boxplot", main = "Zone")
## used log to scale zone variables.
q2 <- qplot(x = Mcrim, y = indus, data = Train, geom = "boxplot", main = "Indus")
q3 <- qplot(x = Mcrim, y = nox, data = Train, geom = "boxplot", main = "nitrogen oxides")
q4 <- qplot(x = Mcrim, y = rm, data = Train, geom = "boxplot", main = "Avg Rooms")
q5 <- qplot(x = Mcrim, y = age, data = Train, geom = "boxplot", main = "Home Units Age")
q6 <- qplot(x = Mcrim, y = dis, data = Train, geom = "boxplot", main = "Distance")
q7 <- qplot(x = Mcrim, y = tax, data = Train, geom = "boxplot", main = "Prop. Tax")
q8 <- qplot(x = Mcrim, y = ptratio, data = Train, geom = "boxplot", main = "Education")
q9 <- qplot(x = Mcrim, y = black, data = Train, geom = "boxplot", main = "Black Pop.")
q10 <- qplot(x = Mcrim, y = lstat, data = Train, geom = "boxplot", main = "Lower Status")
q11 <- qplot(x = Mcrim, y = medv, data = Train, geom = "boxplot", main = "Med Val of Home")
grid.arrange(q1,q2,q3,q4,q5,q6,q7,q8,q9,q10,q11, ncol = 6)## Warning: Removed 280 rows containing non-finite values (stat_boxplot).
Given the graph, five variables, nitroge oxides, Med val Home Value, Distance, Home units Age, lower Status, shows the distinctive characteristics as respect to the degree of crime.
Then, here is the predictive analysis with 3 different models, LDA, GLM, KNN.
ctrl <- trainControl(method = "repeatedcv")
dat <- Train[,-1]
dim(dat)## [1] 380 14findLinearCombos(dat)## $linearCombos
## list()
##
## $remove
## NULLfit.lda <- train(Mcrim ~ . ,trControl = ctrl, data = dat, method = "lda")
fit.lda$finalModel## Call:
## lda(x, grouping = y)
##
## Prior probabilities of groups:
## 0 1
## 0.5 0.5
##
## Group means:
## zn indus chas nox rm age dis
## 0 20.794737 6.993211 0.04736842 0.4718932 6.402337 50.71842 5.07392
## 1 1.168421 15.186211 0.07894737 0.6348263 6.209737 85.11842 2.50559
## rad tax ptratio black lstat medv
## 0 4.178947 305.4895 17.97316 388.6874 9.310579 25.28789
## 1 15.063158 511.7211 19.07526 324.9032 16.026737 19.91684
##
## Coefficients of linear discriminants:
## LD1
## zn -0.003741868
## indus 0.008592637
## chas -0.039542626
## nox 7.869796901
## rm 0.070604292
## age 0.010587569
## dis -0.012376907
## rad 0.076987991
## tax -0.001008256
## ptratio 0.071785521
## black -0.001020335
## lstat 0.016760765
## medv 0.038316240pred.lda <- predict(fit.lda, Test)
fit.glm <- train(Mcrim ~ . ,trControl = ctrl, data = dat, method = "glm", family = "binomial")## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredpred.glm <- predict(fit.glm, Test)
fit.knn <- train(Mcrim ~ . ,trControl = ctrl, data = dat, method = "knn", preProcess = c("center", "scale"), tuneLength = 20)
pred.knn <- predict(fit.knn,Test)In this model, GLM shows the highest accuracy.
cm.lda <- confusionMatrix(Test$Mcrim, pred.lda)
cm.glm <- confusionMatrix(Test$Mcrim, pred.glm)
cm.knn <- confusionMatrix(Test$Mcrim, pred.knn)
D <- data.frame(rbind(cm.lda$byClass, cm.glm$byClass, cm.knn$byClass), row.names = c("LDA", "GLM", "KNN"))
ac <- cbind(cm.lda$overall[1], cm.glm$overall[1], cm.knn$overall[1])
colnames(ac) <- c("LDA", "GLM", "KNN")
D <- rbind(ac,t(D))
kable(D[1:5, ])| LDA | GLM | KNN | |
|---|---|---|---|
| Accuracy | 0.8492063 | 0.8888889 | 0.9047619 |
| Sensitivity | 0.7972973 | 0.8550725 | 0.8695652 |
| Specificity | 0.9230769 | 0.9298246 | 0.9473684 |
| Pos.Pred.Value | 0.9365079 | 0.9365079 | 0.9523810 |
| Neg.Pred.Value | 0.7619048 | 0.8412698 | 0.8571429 |